Question

A solution is prepared by adding 25.0 mL of 1.30 M AlCl3 solution to a flask, and then adding enough water to give a final volume of 200.0 mL. What is the molarity of the solution?

a. 0.260

b. 0.163

c. 6.50

d. 1.24

Answer 1

To calculate the molarity of a diluted solution of AlCl3, use the molarity equation M1V1 = M2V2. The result is a final molarity of about 0.163 M when 25.0 mL of 1.30 M AlCl3 is diluted to 200.0 mL.

Step-by-step explanation:

The question asks about the calculation of molarity after dilution of an AlCl3 solution. To find the molarity after dilution, first note the initial moles of AlCl3 are conserved. You use the formula M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume. Here, M1 = 1.30 M, V1 = 25.0 mL, V2 = 200.0 mL, and we need to find M2.

To solve:

1. Calculate initial moles of AlCl3: (1.30 mol/L) x (25.0 x 10-3L) = 0.0325 mol.
2. Use the formula to get M2: (1.30 M)(25.0 mL) = (M2)(200.0 mL).
3. Solve for M2: M2 = (1.30 M x 25.0 mL) / 200.0 mL = 0.1625 M.

The answer is (b) 0.163 M after rounding to three significant digits.